An ice cube of side `1 cm` is floating at the interface of kerosene and water in a beaker of base area `10 cm^(2)`. The level of kerosene is just covering the top surface of the ice cube.
a. Find the depth of submergence in the kerosene and that in the water.
b. Find the change in the total level of the liquid when the whole ice melts into water.

Answer

`0.1mm`

Explanation

According to the condition of floating
`0.8rho_(w)gh_(k)+rho_(w)gh_(w)=0.9rho_(w)gh`……….i
where `h_(k)` and `h_(w)` are the submerged depth of the ice in the kerosene and water, respectively
`h_(k)+h_(w)=h`…….ii
Solving eqn i and ii we get `h_(k)=0.5cm, h_(2)=0.5cm`
b. `1cm^(3)` ice `overset("melts") (rarr)0.9cm^(3)` water
After melting the ice `0.5cm^(3)` of water will be occupied by the kerosene. Therefore the fall in the level of kerosene is `/_\h_(k)=0.5/A`.
Out of `0.9cm^(3)` of water formed `0.5cm^(3)` of water will be filled in the vacant part of ice and remaining `(0.9-0.5)=0.4cm^(3)` of water will be available to increase the level of water.
Rise in the level of water is `/_\h_(w)=(0.9-0.5)/A=0.4/A`
Hence, the net fall in the overall level is
`/_\h=0.1/A=0.1/10=0.01mm`